Question

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A hungry bear weighing 650 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. 65.0 Goodies The beam is uniform, weighs 210 N and is 5.80 m long; the basket weighs 82.0 N. (a) When the bear is at x = 0.800 m; find the tension in the wire and the components of the force exerted by the wall on the left end of the beam. (b) If the wire can withstand a maximum tension of 900 N, what is the maximum distance that the bear can walk before the wire breaks? O(a) T = 305 N, Fx = 129 N, Fy = 665 N; (b) 5.61 m (a) T = 343 N, Fx = 129 N, Fy = 665 N; (b) 5.61 m (a) T = 305 N, Fx = 129 N, Fy = 665 N; (b) 3.61 m (a) T = 305 N, Fx = 729 N, Fy = 165 N; (b) 5.61 m

Answer 1

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Solution:- let, Wb = 650 Whm = 210 N Wot = 82 N and L= 5.80m So, f.b.d. Let T = tension in the string Trino T od Fyw SS 0 = 65.0 E Tuoso Faw Wom Wb Wot As we that the whole can fel system is in equilibrium. so, 27= as positive and clockwise . Taking anticlockwise direction negative, we have Trino + Wom Who Wbt

Now, Trin OXL woln WomXL WorX L=0 Wo X n + Wam XL/ + Womx 1/2 + WordL Leino 650X 0.8 + 210 X 5.8/2 + 82 x 5.8 5.8 sin 65° 305 N Now, Fxw and Fyw are the components of the force exerted by the wall of the beam, left and The whole system is in equilibriumo ΣΕ, = 0 and E y = 0 Faw Tuos o=0 Fyw - Wo-Won - Wort Tsinos = 0. Few = 305 cos 65° 129N. :. Fyw = 6 50+210+ 82-305 sin 65 665 N

(b) Here. To goON Again wing, E T=0 ha we have Tsin 65 x L - Wox n. Wom WomXL 2 WbtXL = 0 then we have to find the value for a so, и = Trin 65 8L Womx IŲ WbtXL Wo = goo sin 6sx S.- 210x 5.8 2 - 82x s.8 65 0 - 5.61 m So, option option 1st is the correct answer